### Video Transcript

Determine the indefinite integral
of eight times π to the power of three π₯ minus π to the power of two π₯ plus nine
divided by seven π to the power of π₯ with respect to π₯.

In this question, weβre asked to
evaluate the indefinite integral of an exponential expression. And at first, it might look very
difficult to evaluate this integral. For example, we might be
considering a π’-substitution. However, we should always check if
we can simplify our integrand. And we can do this by dividing
every term in the numerator by the denominator, where we remember for each term in
the numerator, we just need to subtract the exponents of π. This gives us the indefinite
integral of eight over seven times π to the power of two π₯ minus one-seventh π to
the power of π₯ plus nine over seven times π to the power of negative π₯ with
respect to π₯.

Now, we can see we know how to
integrate each term in the integrand separately. So, weβre going to split this into
three separate integrals. And we can do this because we
recall the integral of the sum of functions is equal to the sum of their individual
integrals. This gives us the following. We can then simplify each term
separately. Weβll take the constant factor
outside of each integral. This then gives us eight-sevenths
times the indefinite integral of π to the power of two π₯ with respect to π₯ minus
one-seventh the indefinite integral of π to the power of π₯ with respect to π₯ plus
nine-sevenths the integral of π to the power of negative π₯ with respect to π₯.

Weβre now ready to evaluate each of
these indefinite integral separately. We can do this by recalling for any
real constant π not equal to zero, the integral of π to the power of ππ₯ with
respect to π₯ is one over π times π to the power of ππ₯ plus the constant of
integration πΆ. Weβll use this to evaluate each
integral separately. Letβs start with the first
integral, the integral of π to the power of two π₯ with respect to π₯. Our value of π is two. So, we substitute π is two into
our integral rule. We get one-half times π to the
power of two π₯. And we need to multiply this by
eight over seven. And itβs worth noting since weβre
evaluating the sum of multiple indefinite integrals, weβll get a constant of
integration for each of these integrals. We can combine all of these
constants into one constant at the end of our expression. So, we donβt need to worry about
adding πΆ until the end.

Letβs now evaluate our second
integral. We know π to the power of π₯ is
the same as π to the power of one times π₯. So, our value of π will be
one. Of course, this is somewhat not
necessary since we know the integral of π to the power of π₯ with respect to π₯ is
just π to the power of π₯ with respect to π₯. However, if we did substitute π is
equal to one, we would get one over one times π to the power of one π₯, which is
just π to the power of π₯. So, evaluating our second integral
and multiplying by negative one-seventh, we get negative one-seventh π to the power
of π₯.

Finally, letβs evaluate the third
integral. Our value of π is negative
one. Substituting π is negative one
into our integral result gives us one over negative one times π to the power of
negative π₯. We need to multiply this by nine
over seven. And remember, we also need to add a
constant of integration at the end of this expression. We can then simplify this
slightly. One over negative one is just equal
to negative one. So, instead of adding this term, we
can subtract this term. And we can also notice in the first
term, we have a shared factor of two in the numerator and denominator. We can cancel this to get a factor
of four over seven.

This then gives us our final
answer. The indefinite integral of π to
the power of three π₯ minus π to the power two π₯ plus nine over seven π to the π₯
with respect to π₯ is equal to four over seven times π to the power of two π₯ minus
π to the power of π₯ over seven minus nine-sevenths π to the power of negative π₯
plus the constant of integration πΆ.